Difference between revisions of "2001 AIME I Problems/Problem 3"
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== Problem == | == Problem == | ||
− | Find the sum of the | + | Find the sum of the [[root]]s, real and non-real, of the equation <math>x^{2001}+\left(\frac 12-x\right)^{2001}=0</math>, given that there are no multiple roots. |
== Solution == | == Solution == | ||
− | From [[Vieta's formulas]], | + | From [[Vieta's formulas]], in a [[polynomial]] of the form <math>a_nx^n + a_{n-1}x^{n-1} + \cdots + a_0 = 0</math>, then the sum of the roots is <math>\frac{-a_{n-1}}{a_n}</math>. |
− | From the [[Binomial Theorem]], the first term of <math>left(\frac 12-x\right)^{2001}</math> is <math>-x^{2001}</math>, but <math>x^{2001}+-x^{2001}=0</math>, so the | + | From the [[Binomial Theorem]], the first term of <math>\left(\frac 12-x\right)^{2001}</math> is <math>-x^{2001}</math>, but <math>x^{2001}+-x^{2001}=0</math>, so the term with the largest degree is <math>x^{2000}</math>. So we need the coefficient of that term, as well as the coefficient of <math>x^{1999}</math>. |
− | < | + | <cmath>\begin{align*}\binom{2001}{1} \cdot (-x)^{2000} \cdot \left(\frac{1}{2}\right)^1&=\frac{2001x^{2000}}{2}\\ |
+ | \binom{2001}{2} \cdot (-x)^{1999} \cdot \left(\frac{1}{2}\right)^2 &=\frac{-x^{1999}*2001*2000}{8}=-2001 \cdot 250x^{1999} | ||
+ | \end{align*}</cmath> | ||
− | <math>- | + | Applying Vieta's formulas, we find that the sum of the roots is <math>-\frac{-2001 \cdot 250}{\frac{2001}{2}}=250 \cdot 2=\boxed{500}</math>. |
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== See also == | == See also == | ||
{{AIME box|year=2001|n=I|num-b=2|num-a=4}} | {{AIME box|year=2001|n=I|num-b=2|num-a=4}} | ||
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+ | [[Category:Intermediate Algebra Problems]] |
Revision as of 17:34, 11 June 2008
Problem
Find the sum of the roots, real and non-real, of the equation , given that there are no multiple roots.
Solution
From Vieta's formulas, in a polynomial of the form , then the sum of the roots is .
From the Binomial Theorem, the first term of is , but , so the term with the largest degree is . So we need the coefficient of that term, as well as the coefficient of .
Applying Vieta's formulas, we find that the sum of the roots is .
See also
2001 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |